What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$

Question

Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$

(DO NOT use trigonometric methods)

Answer 1

For $a=\frac{1}{\sqrt2}$, $b=\sqrt2$ and $c=\frac{1}{2\sqrt2}$ we'll get a value $\frac{10}{3}$.

We'll prove that it's a maximal value.

Indeed, the condition gives $c(ab+1)=b-a$.

If $ab=-1$ then $a=b$ and $a^2+1=0$, which is impossible.

Thus, $ab\neq-1$, $c=\frac{b-a}{ab+1}$ and we need to prove that $$\frac{2}{1+a^2}-\frac{2}{1+b^2}+\frac{3}{1+\left(\frac{b-a}{1+ab}\right)^2}\leq\frac{10}{3}$$ or $$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq\frac{10}{3}$$ or $$(a^2+4)b^2-18ab+16a^2+1\geq0,$$ for which it's enough to prove that $$81a^2-(a^2+4)(16a^2+1)\leq0,$$ which is $$(2a^2-1)^2\geq0.$$ Done!

I got a value $\frac{10}{3}$ and values $a$, $b$ and $c$ by the following way.

Let $k$ be a maximal value of our expression.

Thus, the following inequality is true for all values of $a$ and $b$. $$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq k$$ or $$(ka^2-3a^2+k-2)b^2-6ab+ka^2+2a^2+k-3\geq0.$$

Thus, $k\geq3$ and since we have a quadratic inequality of $b$, we need that the equality $$9a^2-(ka^2-3a^2+k-2)(ka^2+2a^2+k-3)\leq0$$ or $$(k^2-k-6)a^4+2(k^2-3k-2)a^2+k^2-5k+6\geq0$$ will be true for all value of $a$.

But the last inequality is a quadratic inequality of $a^2$, which says that we need $$(k^2-3k-2)^2-(k^2-k-6)(k^2-5k+6)\leq0$$ or $$3k-10\geq0,$$ which gives a value $\frac{10}{3}$ and from here we can get the case of the equality occurring.

Answer 2

prove that $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}\le \frac{10}{3}$$ and the equal sign holds if $$a=\frac{2}{3}\left(\frac{1}{2\sqrt{2}}-\sqrt{2}\right),b=-\sqrt{2},c=-\frac{1}{2\sqrt{2}}$$